Question
If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what per cent?
✓
Answer
If we consider the Earth to be a perfect sphere, then the acceleration due to gravity at its surface is given by $\text{g}=\text{G}\frac{\text{M}}{\text{R}^2}$ Here, M is the mass of Earth; R is the radius of the Earth and G is universal gravitational constant. If the radius of the earth is decreased by 1%, then the new radius becomes,$\text{R'}=\text{R}-\frac{\text{R}}{100}=\frac{99}{100}\text{R}$
$\Rightarrow\text{R'}=-0.99\text{R}$
New acceleration due to gravity will be given by,$\text{g'}=\text{G}\frac{\text{M}}{\text{R}^2}=\text{G}\frac{\text{M}}{(0.99\text{R})^2}$
$\Rightarrow\text{g'}=1.02\times\Big(\text{G}\frac{\text{M}}{\text{R}^2}\Big)=1.02\text{g}$
Hence, the value of the acceleration due to gravity increases when the radius is decreased. Percentage increase in the acceleration due to gravity is given by,$\frac{\text{g'}-\text{g}}{\text{g}}\times100$
$=\frac{0.02\text{g}}{\text{g}}\times100$
$=2\%$
$\Rightarrow\text{R'}=-0.99\text{R}$
New acceleration due to gravity will be given by,$\text{g'}=\text{G}\frac{\text{M}}{\text{R}^2}=\text{G}\frac{\text{M}}{(0.99\text{R})^2}$
$\Rightarrow\text{g'}=1.02\times\Big(\text{G}\frac{\text{M}}{\text{R}^2}\Big)=1.02\text{g}$
Hence, the value of the acceleration due to gravity increases when the radius is decreased. Percentage increase in the acceleration due to gravity is given by,$\frac{\text{g'}-\text{g}}{\text{g}}\times100$
$=\frac{0.02\text{g}}{\text{g}}\times100$
$=2\%$
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